A note on the gaps between zeros of Epstein's zeta-functions on the ...

A NOTE ON THE GAPS BETWEEN ZEROS OF EPSTEIN’S ZETA-FUNCTIONS ON THE CRITICAL LINE

arXiv:1602.06069v1 [math.NT] 19 Feb 2016

STEPHAN BAIER, SRINIVAS KOTYADA AND USHA KESHAV SANGALE Abstract. It is proved that Epstein’s zeta-function ζQ (s), related to a positive definite integral binary quadratic form, has a zero 1/2+iγ with T ≤ γ ≤ T +T 3/7+ε for sufficiently large positive numbers T . This is an improvement of the result by M. Jutila and K. Srinivas (Bull. London Math. Soc. 37 (2005) 45–53).

To Professor Matti Jutila with deep regards 1. Introduction Let the quadratic form Q(x, y) = ax2 + bxy + cy 2 be positive definite, has integer coefficients and let rQ (n) count the number of solutions of the equation Q(x, y) = n in integers x and y. The Epstein zeta-function associated to Q is denoted by ζQ (s) and is given by the series ζQ (s) =

X

(x,y)∈Z2 −(0,0)

∞ X 1 rQ (n) = Q(x, y)s ns n=1

in the half-plane σ > 1, where (as usual) s = σ + it. Throughout the paper, we shall write ∆ to denote the number ∆ = 4ac − b2 > 0. ζQ (s) has many analytical properties in common with the Riemann zeta-function, ζ(s). For example, it admits analytic continuation into the entire complex plane except for a simple pole at s = 1 with residue 2π∆−1/2 . It satisfies the following functional equation √ !s √ !1−s ∆ ∆ Γ(s)ζQ (s) = Γ(1 − s)ζQ (1 − s) (1.1) 2π 2π The analogue of Hardy’s theorem for ζ(s) also holds true for ζQ (s), i.e., ζQ (s) admits infinitely many zeros on the critical line σ = 1/2. In fact, much more is true. In 1934, Potter and Tichmarsh [3] showed that every interval of the type [T, T + T 1/2+ε ] contains a zero 1/2 + iγ of ζQ (s) for any fixed ε and for all sufficiently large T . Sankaranarayan in 1995 [4] showed that the same result holds true for intervals of the type [T, T + cT 1/2 log T ]. In 2005, Jutila and Srinivas [2] proved that the same is true for intervals of the type [T, T + cT 5/11+ε ], thus surpassing the classical barrier of 1/2 in exponent of T . In this note we improve this result further. More precisely, we prove the following Theorem 1. Let Q be a positive definite binary integral quadratic form. Then for any fixed ε > 0 and T ≥ T (ε, Q), there is a zero 1/2 + iγ of the corresponding Epstein zeta-function ζQ (s) with | γ − T |≤ T 3/7+ε .

(1.2)

Our exposition closely follows the presentation and technical details given in [2], therefore, in this note we indicate only those steps in [2] which enable us to improve the result of Jutila and Srinivas mentioned earlier. The rest of the paper is organized as follows: in section 2 we briefly discuss the main exponential sum of [2], section 3 contains basic results used in the proof of the main theorem and in section 4 we estimate the double exponential sum non-trivially, which leads to the improvement. 2000 Mathematics Subject Classification. 11E45 (primary); 11M41 (secondary). 1

2

STEPHAN BAIER, SRINIVAS KOTYADA AND USHA KESHAV SANGALE

2. The critical exponential sum In order to show that ζQ (s) has a zero 1/2 + iγ with |T − γ| ≤ H, the main point in [2] was to show that the sum X η(n)rQ (n)n−1/2−it (2.1) n

is √ small in a certain √ sense, where η(n) is a smooth weight function supported in the interval [T ∆/2π − K, T ∆/2π + K], t lies close to T , H and K are parameters related by HK = T 1+ε . To estimate the sum (2.1), it was first transformed into another sum (equation (3.6) of [2]) using a transformation formula. By partial summation, this sum got reduced to the estimation of the following term:      X t 1 πQ∗ (x, y) h∆0 ∗ 1/4 −1/4 −1/2 (2.2) + ·φ − e Q (x, y) K N T k 2hk∆0 π 2hk∆0 t ∗ ′ N ≤Q (x,y)≤N

This is equation (4.4) of [2] (see 4 for details). The objective was to show that (2.2) is O(T −ε ). To achieve this, the double exponential sum, appearing above, was estimated non-trivially in one variable using van der Corput’s method and trivial estimate was taken in the other variable. They obtained the bound O(K 11/12 T −1/2 ), which is O(T −ε ), provided K = T 6/11−ε . In the present paper we estimate the above double exponential sum non-trivially in both variables. To show that (2.2) is O(T −ε ), it is now enough to take K = T 4/7−ε and thus the gap H = T 3/7+ε follows. 3. Preliminary lemmas We will use the following well-known lemmas in the proof of our theorem. Lemma 1. (Generalized Weyl differencing) Let a < b be integers, λ be a natural number and ξ(n) be a complex valued function such that ξ(n) = 0 if n 6∈ (a, b]. If H is a positive integer then   X 2 λ|h| X (b − a) + H X 1− ξ(n)ξ(n − λh) (3.1) | ξ(n) | ≤ H H n n |h| 0 such that f (2) (x) ≍ F N −2 , f (3) (x) ≪ F N −3 , and f (4) (x) ≪ F N −4 for x in [a, b]. Let xν be defined by the relation f ′ (xν ) = ν, and let φ(ν) = −f (xν ) + νxν . Then X

a≤n≤b

e(f (n)) =

X e(−φ(ν) − 1/8)

α≤ν≤β

|f ′′ (xν |

1/2

+ O(log(F N −1 + 2) + F −1/2 N ).

(3.2)

Lemma 3. (van der Corput’s bound, Theorem 2.2 of [1]) Suppose that f is a real valued function with two continuous derivatives on [a, b] where a < b are integers. Suppose also that there is some λ > 0 and some α ≥ 1 such that λ ≤ |f ′′ (x)| ≤ αλ

on [a, b]. Then X

a≤n≤b

e(f (n)) ≪ α(b − a)λ1/2 + λ−1/2 .

(3.3)

A NOTE ON THE GAPS BETWEEN ZEROS OF EPSTEIN’S ZETA-FUNCTIONS ON THE CRITICAL LINE 3

4. Estimation of the double exponential sum We start with the integral positive definite quadratic form Q∗ (x, y) ( see equation (3.4) of [2]) : Q∗ (x, y) = a∗ x2 + b∗ xy + c∗ y 2

(4.1)

and let 2

d := (b∗ ) − 4a∗ c∗ < 0.

Note that a∗ > 0, c∗ > 0. As remarked in [2], the form Q∗ (x, y) depends only on Q and k, and |d| ≤ ∆. Our goal is to bound non-trivially the exponential sum ( equation (4.4) of [2] ):   X X  t πQ∗ (x, y) , e Q∗ (x, y) · r + · φ π 2hk∆0 t x

(4.2)

y∈I(x)

where r :=

1 h∆0 − k 2hk∆0

and I(x) := {y ∈ R : N ≤ Q∗ (x, y) ≤ N ′ },

with the convention that the sums are restricted to residue classes mod |d|. Here, 1 ≤ N ≤ N ′ ≤ 2N, N ≪ K, ∆0 |∆, h/k is a rational number satisfying (h∆0 , k) = 1 and

The function φ(x) is defined as

K 1 πK h ≪| √ − |≤ , T T∆ ∆ k p k, h ≍ T /K, t ≍ T.

φ(x) = arsinh(x1/2 ) + (x + x2 )1/2 . In [2], the summation over y− range was evaluated using a classical estimate for exponential sums (Lemma 4.1 of [2]). In the present paper, we explicitly carry out Weyl shift, apply B process and then exploit the cancellations in the sum over x invoking van der Corput’s bound. For the ease of exposition, we restrict ourselves to the case |d| = 1, i.e., x and y vary over all integers. The general case can be treated in a similar way. It is easy to see that I(x) is empty unless x ∈ J , where " √ # √ 2 c∗ N ′ 2 c∗ N ′ J := − p , p |d| |d|

and that

I(x) = I(x) ∪ I ′ (x),

where

and

 v ) s ( u ′ u N |d| |d| N 2 , 2 − I(x) := tmax 0, ∗ − 2 ·x 2 ·x c c∗ (2c∗ ) (2c∗ )  s

Set

I ′ (x) := −

v ) ( u u N′ N |d| |d| tmax 0, 2 2 . − − 2 ·x ,− 2 ·x c∗ c∗ (2c∗ ) (2c∗ )

# √ 2 c∗ N ′ J = {x ∈ J : x ≥ 0} = 0, p . |d| "

4

STEPHAN BAIER, SRINIVAS KOTYADA AND USHA KESHAV SANGALE

In the following, we estimate the partial sum   X X  t πQ∗ (x, y) e Q∗ (x, y) · r + · φ , π 2hk∆0 t x∈J y∈I(x)

where x is positive and y runs over the interval I(x). The remaining three partial sums with a) x ≥ 0 and y ∈ I ′ (x), b) x < 0 and y ∈ I(x), c) x < 0 and y ∈ I ′ (x) can be estimated in a similar way. We start with applying Cauchy-Schwarz, getting   2  ∗ X X t πQ (x, y) e Q∗ (x, y) · r + · φ π 2hk∆0 t x∈J y∈I(x) (4.3)  2  ∗ X X  √ t πQ (x, y) , e Q∗ (x, y) · r + · φ ≪ N· π 2hk∆0 t x∈J y∈I(x) √ √ √ where we use |J| ≪ N . Applying Lemma 1 with λ = 2, and using |J| ≪ N and |I(x)| ≪ N , we have  2  ∗ X X  t πQ (x, y) e Q∗ (x, y) · r + · φ π 2hk∆ t 0 x∈J y∈I(x) √   X X N X |2m| e (fx (y + 2m) − fx (y)) 1− · ≪ · (4.4) M M x∈J y∈Im (x) 0≤|m|≤M √ 3/2 X X X N + NM + N , ≪ · e (f (y + m) − f (y − m)) x x M M 1≤m≤M x∈J y∈I(x)

where M ≤ N is any natural number,

Im (x) := {y ∈ I(x) : y + 2m ∈ I(x)}

and

t fx (y) := Q (x, y) · r + · φ π ∗

We have



πQ∗ (x, y) 2hk∆0 t



.

Q∗ (x, y + m) − Q∗ (x, y − m) = 2m (b∗ x + 2c∗ y) and, using Taylor series expansion,    
πQ∗ (x, y + m) πQ∗ (x, y − m) φ −φ = 2mgx′ (y) + O m3 |gx′′′ (y)| , 2hk∆0 t 2hk∆0 t where

gx (y) := φ Using √ 1 φ (u) = √ + O( u), u ′

′′



φ (u) = O

πQ∗ (x, y) 2hk∆0 t



1 u3/2





.

′′′

and φ (u) = O



1 u5/2



for |u| ≤ 1, we calculate that s     √ ∗ ∗ ∗ (x, y) π (b x + 2c y) 1 πQ π (b∗ x + 2c∗ y) N K 3/2 ′     +O gx (y) = =p · q ∗ +O πQ (x,y) 2hk∆0 t 2hk∆0 t T3 2hk∆0 tQ∗ (x, y) 2hk∆0 t

and

gx′′′ (y)

=O



K 1/2 NT



.

A NOTE ON THE GAPS BETWEEN ZEROS OF EPSTEIN’S ZETA-FUNCTIONS ON THE CRITICAL LINE 5

It follows that X

y∈I(x)

e (fx (y + m) − fx (y − m)) =

X

e (2mFx (y)) + O

y∈I(x)

where ∗

∗

Fx (y) := (b x + 2c y) · We calculate that



mN 3/2 K 3/2 m3 K 1/2 + T2 N 1/2



,

! √ t r+ p . 2πhk∆0 Q∗ (x, y)

√ t|d|x2 , =2c r + √ 2 2πhk∆0 Q∗ (x, y)3/2 √ 3 t|d|x2 (b∗ x + 2c∗ y) K 1/2 Fx′′ (y) = − √ ≍ , N 4 2πhk∆0 Q∗ (x, y)5/2  1/2  K Fx′′′ (y) =O , N 3/2  1/2  K ′′′′ . Fx (y) =O N2 Fx′ (y)

(4.5)

∗

(4.6)

By our assumptions on x and y, we have N ≤ Q∗ (x, y) ≤ N ′ . For fixed x, we also have Q∗ (x, y) ≥

|d| · x2 . 4c∗

Hence a(x) ≤ Fx′ (y) ≤ b(x), where

√ t|d|x2 a(x) := 2c r + √ 2 2πhk∆0 N ′3/2 ∗

and

√ t|d|x2 b(x) :=2c r + √ 2 2πhk∆0 max{N, |d|x2 /(4c∗ )}3/2 ( ) √ 3/2 √ (4c∗ ) t|d|x2 t ∗ √ =2c r + min , p . 2 2πhk∆0 N 3/2 2 2π|d|hk∆0 x ∗

Now Lemma 2 yields X e (2mFx (y)) y∈I(x)

  N 1/2 e (2mFx (yx,m,n ) − nyx,m,n − 1/8) p + O log T + 1/2 1/4 , = m K 2m|Fx′′ (yx,m,n )| 2ma(x)≤n≤2mb(x)

(4.7)

X

where yx,m,n ∈ I(x) is the solution of 2mFx′ (yx,m,n ) = n. We compute that v u √  2/3 u m t|d|x2 t 2 ∗ Fx (yx,m,n ) = −|d|x + 4c √ · 2πhk∆0 (n − 4mc∗ r)

r+



(n − 4mc∗ r) t 2πhk∆0 |d|mx2

and

nyx,m,n =



n  ∗ · −b x + 2c∗

v u u t

−|d|x2 + 4c∗



 √ 2/3 2 m t|d|x  √ . 2πhk∆0 (n − 4mc∗ r)

1/3 !

6

STEPHAN BAIER, SRINIVAS KOTYADA AND USHA KESHAV SANGALE

Putting together gives Gm,n (x) := 2mFx (yx,m,n ) − nyx,m,n v u √ 2/3  ∗ m t|d|x2 b n 1 u t × = ∗ · x − ∗ · −|d|x2 + 4c∗ √ 2c 2c 2πhk∆0 (n − 4mc∗ r)  1/3 ! (n − 4mc∗ r) t ∗ ∗ (n − 4c mr) − 4c m 2πhk∆0 |d|mx2

(4.8)

√  m t|d|x2 (n − 4c∗ mr) b∗ n 2 ∗ √ · x + · −|d|x + 4c 2c∗ 2c∗ |d|x2 2πhk∆0 (n − 4mc∗ r)  3/2 1 m2/3 t1/3 b∗ n = ∗ · x + ∗ · −(n − 4c∗ mr)2/3 |d|1/3 x2/3 + 4c∗ · . 2c 2c (2πhk∆0 )1/3

2/3 !3/2

=

Combining (4.4), (4.5), (4.6), (4.7) and (4.8), we get X X   2  ∗ t πQ (x, y) e Q∗ (x, y) · r + · φ π 2hk∆0 t x∈J y∈I(x) X X X N e (Gm,n (x)) p + ≪ · M |Fx′′ (yx,m,n )| 1≤m≤M x∈J ma(x)≤n≤mb(x)   N2 N2 M N 5/2 K 3/2 3/2 3 1/2 1/2 O MN + +M N K + 1/2 1/4 . + M T2 M K

(4.9)

Let’s first work out what the trivial estimate for the double exponential sum above gives. We have   |J| ≪ N 1/2 and b(x) − a(x) = O K 1/2 N −1/2 .

Together with (4.6), this implies X X X N · M

  e (Gm,n (x)) p = O M 1/2 N 3/2 K 1/4 . 2m|Fx′′ (yx,m,n )| 1≤m≤M x∈J 2ma(x)≤n≤2mb(x)   So choosing M := N 1/6 , using N ≪ K, and taking the square root, we deduce that X X    ∗ t πQ (x, y) e Q∗ (x, y) · r + · φ K 1/4 N −1/4 T −1/2 π 2hk∆0 t x∈J y∈I(x)   11/12 K 25/12 K . + =O T 1/2 T 3/2

(4.10)

(4.11)

So if K := T 6/11−ε , the above is O (T −ε ), as desired. Now we estimate the said double exponential sum non-trivially. First, we re-arrange summations, getting X X X X e (Gm,n (x)) e (Gm,n (x)) p p = , (4.12) ′′ 2m|Fx (yx,m,n )| n∈Jm A (n)≤x≤B (n) 2m|Fx′′ (yx,m,n )| x∈J 2ma(x)≤n≤2mb(x) m

where

Jm



∗

∗

= 4c mr, 4c mr + √

m

√  2c∗ m tN ′ , 2πhk∆0 N 3/2

A NOTE ON THE GAPS BETWEEN ZEROS OF EPSTEIN’S ZETA-FUNCTIONS ON THE CRITICAL LINE 7

Am (n) := (n − 4mc∗ r)

1/2

·

(2πhk∆0 )1/4 N 3/4 (m|d|)1/2 t1/4

and Bm (n) := min

(

3/2

1/4 ′3/4 2(c∗ N ′ )1/2 N (4c∗ ) mt1/2 ∗ 1/2 (2πhk∆0 ) , (n − 4mc r) · , |d|1/2 (m|d|)1/2 t1/4 (2π|d|hk∆0 )1/2 (n − 4mc∗ r)

We note that |Jm | ≪

mK 1/2 N 1/2

and Bm (n) − Am (n) ≪ N 1/2

)

.

(4.13)

and compute that   (n − 4c∗ mr)2/3 |d|1/3 m2/3 t1/3 ∗ 2/3 1/3 2/3 ∗ · (n − 4c mr) |d| x + 4c · × 6c∗ x4/3 (2πhk∆0 )1/3 (4.14) −1/2  m2/3 t1/3 ∗ 2/3 1/3 2/3 ∗ −(n − 4c mr) |d| x + 4c · . (2πhk∆0 )1/3

G′′m,n (x) = −

We have

m2/3 t1/3 m2/3 K 1/3 (n − 4c∗ mr)2/3 |d|1/3 ≍ ≍ N 6c∗ x4/3 (hk∆0 )1/3 N

and (n − 4c∗ mr)2/3 |d|1/3 x2/3 + 4c∗ ·

m2/3 t1/3 m2/3 t1/3 ≍ ≍ m2/3 K 1/3 (2πhk∆0 )1/3 (hk∆0 )1/3

and hence  −1/2 m2/3 t1/3 m4/3 K 2/3 ∗ 2/3 1/3 2/3 ∗ · −(n − 4c mr) |d| x + 4c · ≍ N (2πhk∆0 )1/3 for n and x in the relevant summation intervals. ′ We first assume that n ∈ Jm ⊆ Jm , where √   c∗ m t ′ , Jm := 4c∗ mr, 4c∗ mr + √ 2πhk∆0 N ′ in which case we compute that G′′m,n (x)

−(n − 4c∗ mr)2/3 |d|1/3 x2/3 + 4c∗ ·

(4.15)

m2/3 t1/3 m2/3 t1/3 ≍ ≍ m2/3 K 1/3 (2πhk∆0 )1/3 (hk∆0 )1/3

and hence, using (4.15), mK 1/2 if Am (n) ≤ x ≤ Bm (n). (4.16) N p Now we apply partial summation to remove the factor 1/ Fx′′ (yx,m,n ) and Lemma 3 to get X X e (Gm,n (x)) p 2m|Fx′′ (yx,m,n )| ′ A (n)≤x≤B (n) n∈Jm m m (4.17)   1/2 −1/2 ! mK 1/2 mK 1/2 N 1/2 mK 1/2 1/2 1/2 · · ≪ 1/2 1/4 · mK ≪ mK , + N N m K N 1/2 G′′m,n (x) ≍

where we have used (4.13) and (4.16). ′′ Next, we assume that n ∈ Jm ⊆ Jm , where √ √   2c∗ m tN ′ c∗ m t ∗ ′′ ∗ , 4c mr + √ Jm := 4c mr + √ 2πhk∆0 N ′ 2πhk∆0 N 3/2 and note that in this case x ≍ N 1/2 if Am (n) ≤ x ≤ Bm (n). We further note that ′ ′′ J m = Jm ∪ Jm .

(4.18) (4.19) (4.20)

8

STEPHAN BAIER, SRINIVAS KOTYADA AND USHA KESHAV SANGALE

We write 3/2

Im (n) :=

3/2

(4c∗ ) mt1/2 (4c∗ ) mt1/2 − 1, (2π|d|hk∆0 )1/2 (n − 4mc∗ r) (2π|d|hk∆0 )1/2 (n − 4mc∗ r)

#

and 3/2

3/2

Im (n, δ) :=

(4c∗ ) mt1/2 (4c∗ ) mt1/2 − 2δ, −δ (2π|d|hk∆0 )1/2 (n − 4mc∗ r) (2π|d|hk∆0 )1/2 (n − 4mc∗ r)

#

and set Sm (n) := Im (n) ∩ [Am (n), Bm (n)] and Sm (n, δ) := Im (n, δ) ∩ [Am (n), Bm (n)]. We observe that the interval [Am (n), Bm (n)] can be split into O (log T ) intervals of the form Sm (n) or Sm (n, δ), where √ 1 ≪ δ ≪ N. (4.21) Estimating trivially gives

X

e (Gm,n (x)) p ≪ m1/2 K 1/4 , ′′ (y 2m|F )| x,m,n x (n)

X

′′ n∈Jm x∈Sm

(4.22)

where we have used (4.6) and (4.13). If x ∈ Sm (n, δ), then we compute using (4.18), (4.19) and the mean value theorem that √ 2/3  m2/3 t1/3 δm2/3 K 1/3 m t ∗ 2/3 1/3 2/3 ∗ −1/3 √ −(n − 4c mr) |d| x + 4c · x ≍ ≍ δ · . (2πhk∆0 )1/3 N 1/2 hk∆0 N 1/2 Hence, using (4.15), it follows that mK 1/2 . (4.23) δ 1/2 N 3/4 p Again using partial summation to remove the factor 1/ Fx′′ (yx,m,n ) and Lemma 3, we deduce that X X e (Gm,n (x)) p ′′ 2m|F ′′ x∈S (n,δ) x (yx,m,n )| n∈Jm m (4.24)  1/2  −1/2 ! δmK 1/2 mK 1/2 mK 1/2 mK 1/2 N 1/2 1/2 · · ≪ mK , + ≪ 1/2 1/4 · m K N 1/2 δ 1/2 N 3/4 N 1/2 δ 1/2 N 3/4 G′′m,n (x) ≍

where we have employed |Sm (n, δ)| ≤ δ, (4.13), (4.21) and (4.23). From (4.22) and (4.24), it follows that X X e (Gm,n (x)) p ≪ mK 1/2 log T. (4.25) ′′ (y )| 2m|F x,m,n x n∈J ′′ A (n)≤x≤B (n) m

m

m

Now combining (4.17), (4.20) and (4.25), we get X X e (Gm,n (x)) p ≪ mK 1/2 log T. ′′ (y 2m|F )| x,m,n x n∈Jm A (n)≤x≤B (n) m

(4.26)

m

Combining (4.9), (4.12) and (4.26), we deduce that   2  ∗ X X t πQ (x, y) e Q∗ (x, y) · r + · φ π 2hk∆ t 0 x∈J y∈I(x)   N2 N2 M N 5/2 K 3/2 3/2 1/2 3 1/2 1/2 =O M N + +M N K + 1/2 1/4 + M N K log T . + M T2 M K

(4.27)

A NOTE ON THE GAPS BETWEEN ZEROS OF EPSTEIN’S ZETA-FUNCTIONS ON THE CRITICAL LINE 9

  Choosing M := N 1/4 , using N ≪ K, and taking the square root, it follows that X X    ∗ t πQ (x, y) e Q∗ (x, y) · r + · φ K 1/4 N −1/4 T −1/2 π 2hk∆ t 0 x∈J y∈I(x)   7/8 K 17/8 K . log T + =O T 1/2 T 3/2

This is O(T −ε ), provided that K := T 4/7−ε , which completes the proof.

Acknowledgement. The third author would like to thank the Institute of Mathematical Sciences, Chennai for supporting her visits which enabled her to work on this project. References [1] S. W. Graham and G. Kolesnik, Van der Corput’s Method of Exponential sums, Cambridge University Press, New York, 1991. [2] M. Jutila and K. Srinivas, Gaps between the zeros of Epstein’s zeta-functions on the critical line, Bull. London Math. Soc. 37 (2005) 45-53. [3] H. S. A. Potter and E. C. Titchmarsh, The zeros of Epstein’s zeta-functions, Proc. London Math. Soc. 39(2) (1935) 372–384. [4] A. Sankaranarayanan, Zeros of quadratic zeta-functions on the critical line, Acta Arith. 69 (1995) 21–37. IISER-Thiruvananthapuram, CET Campus, Trivandrum-695016, Kerala, India Institute of Mathematical Sciences, CIT Campus, Taramani, Chennai 600 113, India SRTM E-mail E-mail E-mail

University, Nanded, Maharashtra 431606, India address, Stephan Baier: [email protected] address, Kotyada Srinivas: [email protected] address, Usha Sangale: [email protected]